How do you find the integral $int sqrt(x^3+x^2)(3x^2+2x)dx$ using substitution?

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Answer 1 · 2018-03-26T20:27:06

Let $u = x^3 + x^2$ . Then $du = 3x^2 +2xdx$ and $dx= (du)/(3x^2 + 2x)$ .

$I = int sqrt(u)(3x^2 + 2x)/(3x^2 + 2x) du$

$I = int sqrt(u) du$

$I = 2/3u^(3/2) + C$

$I = 2/3(x^3 + x^2)^(3/2) + C$

Hopefully this helps!

Answer 2 · 2018-03-26T20:50:23

$I=intsqrt(x^3+x^2)(3x^2+2x)dx$ $=int(x^3+x^2)^(1/2)d/(dx)(x^3+x^2)dx=(x^3+x^2)^(1/2+1)/(1/2+1) +c$
$=2/3(x^3+x^2)^(3/2)+c$

Here,

$I=intsqrt(x^3+x^2)(3x^2+2x)dx$

Substituting

$x^3+x^2=u^2=>(3x^2+2x)dx=2udu$

$=>I=intsqrt(u^2)2udu$

$=intu2udu$

$=2intu^2 du$

$=2[u^3/3 ]+c$

$=2/3(u^2)^(3/2)+c$

$=2/3(x^3+x^2)^(3/2)+c$

How do you find the integral int sqrt(x^3+x^2)(3x^2+2x)dxint sqrt(x^3+x^2)(3x^2+2x)dx using substitution?